I'm a second year maths student at the University of Bath. I fell in love with maths after a rocky start with the subject, so believe that I can help people regardless of whether they like maths or not.
I've been tutoring maths locally for a number of years to all ages (youngest was 9, eldest was 82).
In most cases we'll cover what you request. In some cases, however, we'll need to cover some background material prior to the requested topics. This is to aid your understanding and, whilst it might seem tedious at first, missing this knowledge out could result in even more confusion than you had before.
Maths is like a large canvas that's slowly being painted. As you understand a certain area of maths, you paint that area on the canvas. If, however, you try to jump to something without covering the necessary background material, you might end up with something that doesn't fit with the rest of the painting. It is essential to work towards a goal rather than jumping in at the deep end.
I'll use whatever methods work for you, and help you understand it well enough to explain it back to me or your classmates. To quote Peter Singer, "whatever cannot be said clearly is probably not being thought clearly either".
|Maths||A Level||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
The most important thing to remember when differentiating implicitly is that y is a function of x. Rewriting y as y(x) often makes it much clearer. For example, evaluate d/dx (y2): using the aforementioned notation, this becomes d/dx [y(x)]2. By the chain rule, it is easy to see that this is equal to dy/dx * 2y.
Perhaps an easier way of remembering this is to differentiate with respect to y, then multiply by dy/dx. For example, evaluate d/dx(ln(y)): to find the answer, we differentiate ln(y) with respect to y to get 1/y, then multiply this by dy/dx to get dy/dx * 1/y.
The above method works because of the chain rule, which states that df/dx = df/dy * dy/dx. All we are doing is renaming the function as f (in the first example f = y2, in the second example f = ln(y)) and applying this result.see more
Suppose you have 2 functions: f(x) = 3x2, g(x) = log3(x). These are arbitrary, any functions would work. Evaluate f(g(x)): let y = log3(x) ( = g(x) ), then f(g(x)) = f(y) = 3y2 = 3[log3(x)]2.
The part that people tend to find difficult is remembering what it means to apply a function. A simple subsitution makes this much easier. Whilst the above situation makes it seem easy, consider how much more confusing it could be if f(x) = [x7 + 9x5 + e5x + cos(x-1/3)]/[sin(ex/6) + 1729*x], and g(x) was something similarly complicated; a simple substitution can do wonders and will help prevent confusion.see more
The most common way of doing this is called a bubble search (this may not be the technical name, but it's the one I was taught). The method is as follows:
Start by writing the number at the top of your page. Now find 2 numbers which multiply to that number (e.g. starting with 100, I might find 2 and 50). Draw lines down from 100 to the numbers 2 and 50. Check whether or not these numbers are primes (e.g. check whether 2 and 50 are primes - 2 is, 50 isn't). If you have found a prime number, draw a bubble around it. Now do the same process for 50. I find the factors 2 and 25. 2 is prime, 25 isn't. I draw a bubble around 2 and start again from 25. I find the numbers 5 and 5. both are prime so I bubble both.
Now that I've circled both numbers, my search is complete: the prime factors are the numbers in bubbles.
Note that, if neither number is prime, you have to apply the method to both numbers. Because of this, it's often best to find a prime number for one of the 2 numbers.
Some tricks to make this easier are:
If the number is even, use 2 as one of your factors.
If the number ends with a 5 or a 0, use 5 as one of your factors.
If the digits of the number add up to a multiple of 3, use 3 as one of your factors.see more